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Difference between revisions of "2007 AMC 10A Problems/Problem 18"

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The area of <math>\triangle AHG</math> is simply <math>\frac{1}{2} \cdot 4 \cdot 12 = 24</math>, the area of <math>\triangle CGM</math> is <math>\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}</math>, and the area of rectangle <math>ABGH</math> is <math>4 \cdot 12 = 48</math>.
 
The area of <math>\triangle AHG</math> is simply <math>\frac{1}{2} \cdot 4 \cdot 12 = 24</math>, the area of <math>\triangle CGM</math> is <math>\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}</math>, and the area of rectangle <math>ABGH</math> is <math>4 \cdot 12 = 48</math>.
  
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Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}\ \text{(C)}</math>.
+
Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - \left(24 + \frac{32}{5}\right) = \boxed{\frac{88}{5}}\ \text{(C)}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>.
 
The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>.
  
−
Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up  
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Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM_1 = \frac 35 \cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, <math>MM_2</math> must be <math>\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}</math>, by the harmonic mean. Thus, <math>MM_1</math> must be <math>\frac{36}{5}</math>.) 
  
 
Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 =  \boxed{\frac{88}5}</math>.
 
Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 =  \boxed{\frac{88}5}</math>.
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 +
== Solution 3 ==
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<asy>
 +
unitsize(13mm);
 +
defaultpen(linewidth(.8pt)+fontsize(10pt));
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dotfactor=4;
 +
 +
pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2);
 +
pair M=intersectionpoints(A--G,H--C)[0];
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 +
draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle);
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draw(A--G);
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draw(H--C);
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dot(M);
 +
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label("$A$",A,NW);
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label("$B$",B,NE);
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label("$C$",C,NE);
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label("$D$",D,NE);
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label("$E$",Ep,SE);
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label("$F$",F,SE);
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label("$G$",G,SE);
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label("$H$",H,SW);
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label("$I$",I,SW);
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label("$J$",J,SW);
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label("$K$",K,NW);
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label("$L$",L,NW);
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label("$M$",M,W);
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</asy>
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 +
We can use coordinates to solve this. Let <math>H=(0,0).</math> Thus, we have <math>A=(0,12), C=(4,8), G=(4,0).</math> Therefore, <math>AG</math> has equation <math>-3x+12=y</math> and <math>HC</math> has equation <math>2x=y.</math> Solving, we have <math>M=(\frac {12}{5},\frac {24}{5}).</math> Using the [[Shoelace Theorem]] (or you could connect <math>LC</math> and solve for the resulting triangle + trapezoid areas), we find <math>[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.</math>
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 +
~(minor edits by Arcticturn)
  
 
==See also==
 
==See also==

Latest revision as of 23:17, 19 May 2023

Problem

Consider the $12$-sided polygon $ABCDEFGHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$. What is the area of quadrilateral $ABCM$?

[asy] unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,NE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W); [/asy]

$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$

Solution

Solution 1

We can obtain the solution by calculating the area of rectangle $ABGH$ minus the combined area of triangles $\triangle AHG$ and $\triangle CGM$.

We know that triangles $\triangle AMH$ and $\triangle GMC$ are similar because $\overline{AH} \parallel \overline{CG}$. Also, since $\frac{AH}{CG} = \frac{3}{2}$, the ratio of the distance from $M$ to $\overline{AH}$ to the distance from $M$ to $\overline{CG}$ is also $\frac{3}{2}$. Solving with the fact that the distance from $\overline{AH}$ to $\overline{CG}$ is 4, we see that the distance from $M$ to $\overline{CG}$ is $\frac{8}{5}$.

The area of $\triangle AHG$ is simply $\frac{1}{2} \cdot 4 \cdot 12 = 24$, the area of $\triangle CGM$ is $\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}$, and the area of rectangle $ABGH$ is $4 \cdot 12 = 48$.

Taking the area of rectangle $ABGH$ and subtracting the combined area of $\triangle AHG$ and $\triangle CGM$ yields $48 - \left(24 + \frac{32}{5}\right) = \boxed{\frac{88}{5}}\ \text{(C)}$.

Solution 2

[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W); label("$N$",Z,NE); [/asy]

Extend $AB$ and $CH$ and call their intersection $N$.

The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\frac{2\cdot 4}2 = 4$.

The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$.

Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \frac 35 \cdot 12 = \frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\frac{36}{5}$.)

Then the area of $AMN$ is $\frac 12 \cdot AN \cdot MM_1 = \frac{108}5$, and the area of $ABCM$ can be obtained by subtracting the area of $BCN$, which is $4$. Hence the answer is $\frac{108}5 - 4 = \boxed{\frac{88}5}$.

Solution 3

[asy] unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,NE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W); [/asy]

We can use coordinates to solve this. Let $H=(0,0).$ Thus, we have $A=(0,12), C=(4,8), G=(4,0).$ Therefore, $AG$ has equation $-3x+12=y$ and $HC$ has equation $2x=y.$ Solving, we have $M=(\frac {12}{5},\frac {24}{5}).$ Using the Shoelace Theorem (or you could connect $LC$ and solve for the resulting triangle + trapezoid areas), we find $[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.$

~(minor edits by Arcticturn)

See also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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